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3 years ago

To produce a magnetic field, What does an electromagnet require?

Physics

2 answers:

kodGreya [7K]3 years ago

8 0

Selection D makes most sense

AURORKA [14]3 years ago

7 0

Answer:d

Explanation:

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The substances listed on the left side of a chemical equation are the _____.

aivan3 [116]

<span>Reactants
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3 years ago

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A bag of sugar weights 20 N on the earths surface. If you double the distance from the center of the earth, the bag now weighs w

elena-s [515]

I have no idea that’s a hard one

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3 years ago

The conduction velocity of an axon is determined by:

Marina86 [1]

The conduction velocity of an axon is determined by myelin sheath

thickness and internode distance.

Axon are structures in the neuron which is involved in the conduction of

impulses away from the cell body. Axons which have myelin sheath conduct

impulses faster than those without it.

Axons which have thicker myelin sheath and longer internode distance will

increase the conduction velocity of an axon and vice versa.

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3 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

1 year ago

Arbeitsauftrag 2

kramer

Explanation:

<em>The height of the pendulum is measured from the lowest point it reaches (point 3). </em>

At 1, the kinetic energy of the pendulum is zero (because it is not moving), and it has maximum potential energy.

At 2, the pendulum has both kinetic and potential energy, and how much of each it has depends on its height—smaller the height greater the kinetic energy and lower the potential energy.

At 3, the height is zero; therefore, the pendulum has no potential energy, and has maximum kinetic energy.

At 4, the pendulum again gains potential energy as it climbs back up, Again how much of each forms of energy it has depends on its height.

At 5, the maximum height is reached again; therefore, the pendulum has maximum potential energy and no kinetic energy.

Hope this helps :)

8 0

3 years ago